Overview of normed linear spaces

Transcription

1 20

2 Chapter 2 Overview of normed linear spaces Starting from this chapter, we begin examining linear spaces with at least one extra structure (topology or geometry). We assume linearity; this is a natural feature of functional spaces. In this chapter, we start with spaces whose geometric and linear structures are compatible. More precisely, we assume that the metric is translation invariant d(x, y) = d(x + z, y + z) and homogeneous d(λx, λy) = λ d(x, y). We then define x = d(x, 0) the norm of x, and it is clear that. satisfies the following three properties (below x, y X and λ C): x 0 and equality holds iff x = 0. x + y x + y. λx = λ x. A linear space equiped with such a norm is called a normed linear space (NLS). Conversely, if such a norm exists we could always define a metric d(x, y) = x y that is translation invariant and homogeneous. If the norm topology is separable then we say the NLS is separable. A Banach space is a complete normed linear space, i.e. the induced metric space is complete. A Hilbert space has additional structures, where we could talk about angle through the scalar product. We ll discuss these spaces further in separate chapters. We recall that an incomplete metric space could be completed, and this applies to normed linear spaces too: the metric remains invariant under dilation and translation in the completed space thus it remains a norm. Some examples: 1. Let (X, µ) be a measure space. Let 1 p. Then L p (X, µ) are normed linear spaces, with f p = ( f(x) p dµ) 1/p X 21

3 22 CHAPTER 2. OVERVIEW OF NORMED LINEAR SPACES In fact they are complete, this is a theorem of Riesz-Fisher. To see this, suffices to show that if {f n } has n f n p < then n f n converges. 1 Now by the triangle inequality ( n f n ) p dµ <, thus h(x) = n f n(x) is finite for µ- almost every x. Thus n f(x) converges to some g(x) measurable, clearly g p h p < so g L p. Finally, ( n f n) g p converges pointwise a.e. to 0 and dominated by (h(x) + g(x) ) p which are integrable, thus by dominated convergence ( n f n) g p dµ 0, thus f n g in L p. On the other hand, for p < weak L p (X, µ) is not a normed linear space, since the triangle inequality fails f p, = sup λµ({x X : f > λ}) 1/p λ>0 However, we still say the weak L p norm in practice. Here is a counter example to the triangle inequality: use f(x) = x1 0 x 1 and g(x) = (1 x)1 0 x If X is compact Hausdorff then C(X) is a complete normed linear space if we use the sup norm f = sup f(x) x X If X is not compact then the sup is not necessarily finite and so this is not even a norm. For locally compact (Hausdorff) spaces we could instead look at C c (X) consisting of compactly supported continuous functions, and this space is a normed linear space. However, C c (X) is not complete in the noncompact case, infact its completion is C 0 (X) the space of continuous functions on X that vanish at. (These are functions f C(X) such that for every M > 0 the set { f M} is compact.) To see this, we first show that C 0 (X) is complete under the uniform norm. One way to show this is to use one-point compactification. Alternatively, consider a Cauchy sequence (f n ) in C 0 (X), then for each x X f n (x) is a Cauchy sequence of complex numbers, thus it converges pointwise to some f(x). Furthermore on any compact subset of X the uniform convergence f n (x) f(x) implies continuity of f. Since X is locally compact and continuity is a local property, it follows that f C(X), then using f n C 0 it is not hard to see that f C 0. (One way to check this is to use the net convergence characterization of continuity). Now, given a function f C 0 (X) we could approximate it by a convergent sequence of compactly supported continuous functions on X using Urysohn s lemma: let K n = { f 1/n} which is compact and sits inside U n = { f < 1/(n 1)} an open set. Then by Urysohn s lemma we may find φ n a bump function that vanishes outside U n but equals 1 on K n, it is clear that φ n f f in the uniform metric. 3. L 2 -Sobolev spaces on R. If k 0 integer we could define H k := {f L 2 : f,..., f (k) L 2 } 1 For normed linear spaces, completeness is equivalent to every absolutely summable sequence is summable.

4 2.1. LINEAR FUNCTIONALS AND DUAL SPACES 23 be the space of L 2 functions whose first k derivatives exists almost everywhere and are L 2 integrable. H k is complete and we may alternatively define H k as the completion of the space of locally compacted C functions on R under the norm f H k = f f (k) 2 Using the Fourier transform, we could generalize this to allow for k fractional and even negative, and we could also use L p instead of L 2. We ll revisit this in the future if time permits. 4. If Y is a closed subspace of X then the quotient space X/Y (the space of equivalent classes where x 1 x 2 if x 1 x 2 Y ) is a normed linear space with norm [x] = inf x y y Y An equivalent definition is [x] = inf z [x] z. Note that closedness of Y is essential here to ensure that [x] = 0 iff x Y. 5. If Y is a subspace of X then the closure Y of Y (with respect to the norm topology) is another subspace of X. If this closure is the same as X then we say that Y is dense in X. Note that this closure is not necessarily complete and Y is not the same as the completion of Y under the norm. For instance we could take X = Y incomplete NLS, then the closure of Y under the norm topology of X is the same as X, still incomplete. 2.1 Linear functionals and dual spaces There are two basic notions of dual spaces: algebraic dual, which consists of all linear maps (aka linear functionals) l : X R (or C); and continuous dual, which consists of all continuous linear functionals. We will be only interested in continous dual spaces, which will be implicitly understood whenever we refer to dual spaces. Note that we could define continuous dual spaces even if X is only a topological linear space, without any norm Linear functionals For a normed linear space, we could talk about boundness of a linear functional. We say that a linear functional l is bounded if there exists C > 0 such that for every x X it holds that l(x) C x Equivalently, this means sup x X: x 1 l(x) < (equivalently sup x X: x =1 l(x) < ) which sometimes is used as the definition. Note that this fact also holds for linear transformations from X to a Banach spaces.

5 24 CHAPTER 2. OVERVIEW OF NORMED LINEAR SPACES Theorem 5. On normed linear spaces, a linear functional is continuous iff it is bounded. Proof. In one direction, clearly being bounded implies being continuous. For the other direction, we d like to show that the image of the closed unit ball under l is a bounded set if l is linear continuous. Note that l maps compact sets to compact sets, but unfortunately as we ll see the closed unit ball is not compact if X is infinite dimensional. So one has to exploit linearity of l. Assume towards a contradiction that x n is a sequence of unit vector s.t. l(x n ) > n. Then x n /n converges to 0 in the norm, but l(x n /n) > 1 which violates continuity of l. One of the most frequently/implicitly used theorems for bounded linear functionals on normed linear space is the socalled B.L.T. (bounded linear transformation) theorem. The theorem applies even for bounded linear maps from X to another Banach space. Theorem 6. Let D be a dense subspace of the normed linear space X. Let l : D Y where Y is a Banach space, and for some C > 0 it holds for all x D that l(x) Y C x X Then l has an unique extension to a bounded linear map from X to Y and satisfies the above estimate for all x X. Proof. It is not hard to see that if such l exists it has to be unique. To define l, fix x X. Since D is dense in X there exists a sequence (x n ) in D that converges to x. We then define lx = lim lx n. Note that this limit exists because lx n is a Cauchy sequence in Y which is a complete space. One could easily show that the value of lx does not depend on the choice of the sequence x n. Linearity and boundedness could be easily checked. While working with singular integral operators on L p spaces we typically invoke the above theorem implicitly: these operators are explicitly defined only for a nicer dense subset of functions (say sufficiently smooth and with sufficient decay), and so the theorem says that as long as we could bound the operators on these dense subspace we could extend the operator to all of the corresponding L p and get the same bound. Example: the Hilbert transform Hf(x) = p.v. 1f(x y)dy is defined for smooth R y compactly supported functions on R which is dense in L p. It turns out that Hf p C f p, 1 < p <, thus H extends to a bounded maps on L p Dual spaces For a bounded linear functional we could define l = sup x X: x =1 l(x) and this defines a norm on the dual space of X. By definition l(x) x l. Theorem 7. The dual space of a normed linear space is a Banach space. Proof. This is actually a special case of a more general fact: the space B(X, Y ) of bounded operators from normed linear space X to a Banach space Y T x Y T x X

6 2.1. LINEAR FUNCTIONALS AND DUAL SPACES 25 is in turn another Banach space. (For us Y = R (or C) with the distance norm). It is not hard to see that this is a normed linear space, the main thing is to show completeness. Given any Cauchy sequence (T n ) in B(X, Y ) it is not hard to see that sup n T n <. Now for any x X the sequence T n x is a Cauchy sequence in Y therefore it converges (thanks to completness of Y ) in Y, and we let T x to be this limit. It is clear that T is linear and bounded T sup n T n <. It remains to show that lim n T n T = 0. Examples: 1. The dual space of a Hilbert space is itself. 2. If 1 < p < then the dual space of L p (X, µ) is L q (X, µ) where q is the conjugate exponent 1/p + 1/q = 1. In particular, f L p (X,µ) = sup fgdµ g: g q=1 If µ is a σ-finite measure(namely one could break the space into countably many subsets where µ is finite) then the dual space of L 1 (X, µ) is L (X, µ). This may not be true without the σ-finite assumption. ON the other hand, the dual space of L (X, µ) generally speaking not L 1 (X, µ) (but there are examples when they are, say when X is a finite set with the counting measure). 3. If X is a locally compact Hausdorff space, then the dual of C 0 (X) is the space of regular Borel measures with finite total mass. This result is one of Riesz s representation theorems and we will prove them later in the course Reflexivity If X is a normed linear space we let X denote its dual and X denote the dual of X. Definition 3 (Reflexive spaces). We say that X is reflexive if X = X (up to isomorphism). In particular, a reflexive space has to be a Banach space to begin with, but certainly not all Banach spaces are reflexive. The interest in reflexive spaces is natural, since we always can isometricaly embed X into X. To see this, fix x X. Let. and. be the norms on X and X respectively. We could map x X to the following linear map x on X : x(l) := l(x), l X. Note that x is bounded on X, since by definition x(l) x l. Thus x x. On the other hand, using the Hahn Banach theorem it follows that we could find l X such that l(x) = x and l(y) y for all y X. It follows that l 1, therefore x(l) = x x l, and consequently x x. Thus x = x and the embedding x x is an isometric embedding of X into X. As we ll see, being reflexive helps in many situation; we ll explore that gradually.

7 26 CHAPTER 2. OVERVIEW OF NORMED LINEAR SPACES Examples: 1. Hilbert spaces are reflexive. 2. L p are reflexive if 1 < p <. As discussed above, generally speaking both L 1 and L are not reflexive. For examples they are not reflexive when the underlying space is R n with Lebesgue measure or Z with counting measure. 3. A closed linear subspace of a reflexive space is also reflexive. 4. C[ 1, 1] is not reflexive. Note that this space is separable but its dual is not (since the dual of this space contains in particular all Borel measures). As we ll see later, for a reflexive space its dual space has to be separable too. One could prove this directly w/o using separability. 2.2 (Non)compactness of the unit ball One of the key themes in analysis is existence of limit of a sequence (in some functional spaces) that are bounded uniformly in norm. By rescaling if necessary we may assume the bound is 1. Sometimes, it suffices to get a convergent subsequence. In other words we are interested in the sequential compactness of the closed unit ball. Since NLS is in particular a metric space, sequential compactness = compactness. So the question is about compactness of the closed unitball in the norm topology. For Euclidean spaces it is clear that for finite dimensional setting the closed unit ball is compact. For infinite dimensional setting, the answer is no, for instance consider the sequence a n = (0,..., 0, 1, 0,... ) (the nth coordinate is 1). If there is a limit x = (x 1, x 2,... ) for some subsequence a nk, i.e. a nk x 0 as k. Then clearly x = 1. On the other hand, given any j by taking k large it is clear that 0 x j a nk x 0, thus x 1 = x 2 = = 0, which is a contradition. This suggests that the answer should be similar for a general normed linear space: positive for spaces with finite dimensions and negative in the infinite dimensional setting (whether the space is complete or not). In the finite dimension case, this follows from the fact that in finite dimensional linear spaces all norms are equivalent. In other words, if. 1 and. 2 are two norms then there exists C 1, C 2 > 0 such that 1 C 1 x 1 x 2 C 2 x 1 for all x X. To see this, for simplicity consider linear spaces over R, and assume that the given space is spanned by x 1,..., x m, it suffices to show the conclusion for. 2 being the Euclidean norm a 1 x a m x m 2 = a a 2 m the Euclidean norm. Then by the triangle inequality a 1 x a m x m 1 ( a a m ) max x j 1 m a a 2 m max x j 1 j j so we could take C 1 = m max j x j 1. To show existence of C 2, note that the identity map l : (X,. 2 ) (X,. 1 ) (i.e. lx = x) is continuous since it is Lipschitz. The unit ball

8 2.2. (NON)COMPACTNESS OF THE UNIT BALL 27 in (X,. 2 ) is compact, thus its image under this continous function is also compact, thus bounded in. 1, therefore for some C 2 > 0 we have sup x 1 C 2 the sup is over x with x 2 = 1, which is equivalent to the desired estimate. In the infinite dimensional case, this is a theorem of Riesz. Theorem 8 (Riesz). If X is an infinite dimensional NLS then the closed unit ball is not compact with respect to the norm topology. Proof. Our plan is to find a sequence of elements of unit vectors in X, x 1, x 2,..., such that the distance x i x j between any two elments of the sequence is uniformly larger than 0, for instance x i x j 1 for all i j. If that could be constructed it is clear that 10 no subsequence of (x n ) is Cauchy, thus no subsequence of this sequence is convergent and therefore the closed unit ball is not (sequentially) compact. To construct this sequence it suffices to show that if Y is a closed proper subspace of X then one could find x X \ Y with x = 1 such that dist(x, Y ) 1. Once this is proved 10 we could start with any point x 1 of unit length and let Y be spaned by x 1 (which is closed) and then select x 2 of unit length as above (in particular x 2 x 1 1 ), then reset Y to be 10 spaned by x 1, x 2 (a closed subspace because it has finite dimension) and select x 3, etc. Thus we only to show existence of x X of unit norm satisfying dist(x, Y ) 1 whenever 10 Y is a closed proper subspace of X. Let z X \ Y and let d := dist(z, Y ) > 0. (if d = 0 then the closedness of Y would imply that z Y, contradiction.) Now for some y Y we have z y > d/10. Let x = z y, it follows that dist(x, Y ) = dist(z, Y ) = d, thus x dist(x, Y )/10 now we just rescale x so that it has length 1. Because of this result, there is a natural question of determining a topology on X so that the closed unit ball is compact with respect to this topology. One wants a weaker topology (with fewer open sets, too many open sets is probably the reason why the ball is not compact). There is a natural notion of weak-topology that is weaker than the norm topology. Here we wants just enough open sets so that all bounded linear functionals are continuous. Theorem 9. The closed unit ball is compact with respect to the weak-topology if and only if X is reflexive. Note that compactness of the closed unit ball is compact in this weak topology does not imply local compactness of X. We ll prove this result later when discussing topologies on Banach spaces; as we ll see this theorem is a consequence of the Banach Alaoglu theorem.

9 28 CHAPTER 2. OVERVIEW OF NORMED LINEAR SPACES Exercises: 1. Prove that for a normed linear space, completeness of the norm is equivalent to every absolutely summable sequence is summable. 2. Prove that for every x X a normed linear space it holds that x = sup l X : l =1 l(x). 3. Prove that if K is compact Hausdorff then C(K) is complete with respect to the sup norm. Use this to complete the proof that C 0 (X) is complete if X is locally compact Hausdorff without appealing to the one-point compactification trick. 4. Let S be a subset of X a normed linear space, and let Y be the linear span of S (consisting of all finite linear combination of elements of S). Let L X to be the set of all l X that vanishes on S. Prove that z Y the closure of Y if any only if l(z) = 0 for every l L. (Hint: one direction should be easier, for the other direction you should use the BLT theorem somewhere.) 5. Prove that all closed linear subspaces of a reflexive (Banach) space are reflexive. [Hint. Use problem 4 at some point.]